A100398 -id:A100398 - cp's OEIS frontend

A055573 Number of terms in simple continued fraction for n-th harmonic number H_n = Sum_{k=1..n} (1/k).

1, 2, 3, 2, 5, 4, 6, 7, 10, 8, 7, 10, 15, 9, 9, 17, 18, 11, 20, 16, 18, 18, 23, 19, 24, 25, 24, 26, 29, 21, 24, 23, 26, 25, 32, 34, 33, 26, 24, 31, 32, 31, 36, 36, 39, 32, 34, 42, 47, 44, 46, 35, 40, 48, 43, 47, 59, 50, 49, 39, 50, 66, 54, 44, 54, 49, 41, 64, 47, 46, 54, 71, 72

Offset: 1

Leroy Quet, Jul 10 2000

nonn

By "simple continued fraction" is meant a continued fraction whose terms are positive integers and the final term is >= 2.

Does any number appear infinitely often in this sequence?

			Sum_{k=1 to 3} [1/k] = 11/6 = 1 + 1/(1 + 1/5), so the 3rd term is 3 because the simple continued fraction for the 3rd harmonic number has 3 terms.

Steven R. Finch, Mathematical Constants, Cambridge, 2003, pp. 156

G. C. Greubel, Table of n, a(n) for n = 1..10000 (terms 1..500 from M. F. Hasler)
Eric Weisstein's World of Mathematics, Harmonic Number
Eric Weisstein's World of Mathematics, Continued Fraction
G. Xiao, Contfrac server, To evaluate H(m) and display its continued fraction expansion, operate on "sum(n=1, m, 1/n)"

m-th harmonic number H(m) = A001008(m)/A002805(m).
Cf. A058027, A100398, A110020, A112286, A112287.
Cf. A139001 (partial sums).

Table[ Length[ ContinuedFraction[ HarmonicNumber[n]]], {n, 1, 75}] (* Robert G. Wilson v, Dec 22 2003 *)

c=0;h=0;for(n=1,500,write("projects/b055573.txt",c++," ",#contfrac(h+=1/n))) \\ M. F. Hasler, May 31 2008

from sympy import harmonic
from sympy.ntheory.continued_fraction import continued_fraction
def A055573(n): return len(continued_fraction(harmonic(n))) # Chai Wah Wu, Jun 27 2024

It appears that lim n -> infinity a(n)/n = C = 0.84... - Benoit Cloitre, May 04 2002

Conjecture: limit n -> infinity a(n)/n = 12*log(2)/Pi^2 = 0.84..... = A089729 Levy's constant. - Benoit Cloitre, Jan 17 2004

A058027 Sum of terms of continued fraction for n-th harmonic number, 1 + 1/2 + 1/3 + ... + 1/n.

Original entry on oeis.org

1, 3, 7, 14, 15, 10, 16, 19, 26, 35, 72, 41, 38, 79, 83, 42, 59, 143, 68, 61, 70, 51, 50, 78, 74, 82, 130, 113, 111, 315, 235, 1190, 211, 407, 112, 122, 142, 246, 693, 133, 138, 162, 1904, 243, 170, 539, 363, 210, 197, 518, 275, 502, 527, 316, 1729, 224, 228, 909

Offset: 1

Leroy Quet, Nov 15 2000

Is anything known about the asymptotics of this sequence?

Should be asymptotic to D*n^(3/2) with D=0.4.... - Benoit Cloitre, Dec 23 2003

			1 + 1/2 +1/3 = 11/6 = 1 + 1/(1 + 1/5). So sum of terms of continued fraction is 1 + 1 + 5 = 7.

G. C. Greubel, Table of n, a(n) for n = 1..1000
Eric Weisstein's World of Mathematics, Harmonic Number
Eric Weisstein's World of Mathematics, Continued Fraction

m-th harmonic number H(m) = A001008(m)/A002805(m).

Cf. A055573, A100398, A110020, A112286, A112287.

Table[Plus @@ ContinuedFraction[HarmonicNumber[n]], {n, 60}] (* Ray Chandler, Sep 17 2005 *)

a(n) = vecsum(contfrac(sum(k=1, n, 1/k))); \\ Michel Marcus, Mar 23 2017

A110020 Final term of the simple continued fraction for H(n), where H(n) = Sum_{k=1..n} 1/k.

Original entry on oeis.org

1, 2, 5, 12, 8, 2, 5, 7, 3, 2, 5, 4, 6, 13, 7, 2, 11, 15, 6, 2, 36, 13, 2, 6, 3, 7, 3, 4, 2, 9, 4, 2, 2, 2, 2, 2, 6, 5, 2, 3, 2, 2, 2, 2, 11, 3, 59, 8, 2, 4, 104, 103, 5, 6, 2, 2, 2, 59, 2, 2, 3, 9, 20, 4, 2, 3, 4, 3, 4, 2, 2, 2, 2, 2, 2, 4, 3, 4, 2, 3, 2, 37, 2, 49, 6, 2, 6, 10, 2, 4, 8, 15, 2, 2, 23, 2

Offset: 1

Leroy Quet, Sep 03 2005

			H(5) = 137/60 = 2 + 1/(3 + 1/(1 + 1/(1 + 1/8))); a(5) is the final term, 8.

G. C. Greubel, Table of n, a(n) for n = 1..1000
Eric Weisstein's World of Mathematics, Harmonic Number
Eric Weisstein's World of Mathematics, Continued Fraction

m-th harmonic number H(m) = A001008(m)/A002805(m).

Cf. A055573, A058027, A100398, A112286, A112287.

Table[Last[ContinuedFraction[HarmonicNumber[n]]], {n, 100}] (* Ray Chandler, Sep 17 2005 *)

Extended by Ray Chandler, Sep 17 2005

A112286 a(n) = numerator of sum of reciprocals of the terms of the continued fraction for H(n) = Sum_{k=1..n} 1/k.

Original entry on oeis.org

1, 3, 11, 7, 71, 7, 17, 152, 2699, 701, 691, 248, 133, 137, 61933, 809, 20705, 64896, 3587, 17449, 445, 61897, 208, 20663, 1163, 982, 27281, 1871, 2466139, 44339, 21293609, 13417971, 6229, 54238033, 99737, 3585191, 33583, 40756259, 5956441

Offset: 1

Leroy Quet, Sep 01 2005

			1 +1/2 +1/3 +1/4 +1/5 +1/6 = 49/20 = 2 + 1/(2 + 1/(4 + 1/2)).
So a(6) is 7, the numerator of 7/4 = 1/2 + 1/2 + 1/4 + 1/2.

G. C. Greubel, Table of n, a(n) for n = 1..1000
Eric Weisstein's World of Mathematics, Harmonic Number
Eric Weisstein's World of Mathematics, Continued Fraction

m-th harmonic number H(m) = A001008(m)/A002805(m).

Cf. A055573, A058027, A100398, A110020, A112287.

f[n_] := Plus @@ (1/# &) /@ ContinuedFraction[Sum[1/k, {k, n}]]; Table[Numerator[f[n]], {n, 40}] (* Ray Chandler, Sep 06 2005 *)

Extended by Hans Havermann and Ray Chandler, Sep 06 2005

A112287 a(n) = denominator of sum of reciprocals of the terms of the continued fraction for H(n) = Sum_{k=1..n} 1/k.

Original entry on oeis.org

1, 2, 5, 12, 24, 4, 5, 35, 420, 156, 300, 45, 15, 39, 15351, 72, 1848, 10675, 300, 2142, 36, 5460, 15, 1870, 90, 63, 2040, 120, 138600, 3960, 1750320, 1324895, 440, 3945480, 5220, 158340, 1680, 3341100, 498960, 48048, 1260, 69264, 1510, 1168200, 568260

Offset: 1

Leroy Quet, Sep 01 2005

			1 +1/2 +1/3 +1/4 +1/5 +1/6 = 49/20 = 2 + 1/(2 + 1/(4 + 1/2)).
So a(6) is 4, the denominator of 7/4 = 1/2 + 1/2 + 1/4 + 1/2.

G. C. Greubel, Table of n, a(n) for n = 1..1000
Eric Weisstein's World of Mathematics, Harmonic Number
Eric Weisstein's World of Mathematics, Continued Fraction

m-th harmonic number H(m) = A001008(m)/A002805(m).

Cf. A055573, A058027, A100398, A110020, A112286.

f[n_] := Plus @@ (1/# &) /@ ContinuedFraction[Sum[1/k, {k, n}]]; Table[Denominator[f[n]], {n, 45}] (* Ray Chandler, Sep 06 2005 *)

Extended by Hans Havermann and Ray Chandler, Sep 06 2005

A139001 Partial sums of A055573 = number of terms in continued fraction of H(n)=sum(1/k,k=1..n).

Original entry on oeis.org

1, 3, 6, 8, 13, 17, 23, 30, 40, 48, 55, 65, 80, 89, 98, 115, 133, 144, 164, 180, 198, 216, 239, 258, 282, 307, 331, 357, 386, 407, 431, 454, 480, 505, 537, 571, 604, 630, 654, 685, 717, 748, 784, 820, 859, 891, 925, 967, 1014, 1058, 1104, 1139, 1179, 1227, 1270

Offset: 1

M. F. Hasler, May 31 2008

nonn

Sequence A100398 holds the array having as n-th row the continued frac. of H(n); a(n) is the last term of the n-th row and accordingly, a(n-1)+1 is the index where the n-th row starts.

M. F. Hasler, Table of n, a(n) for n = 1..500

Cf. A001008, A002805, A055573, A058027, A100398, A110020, A112286, A112287.

h=s=0;vector(100,n,s+=#contfrac(h+=1/n))

a(n) = sum_{k=1..n} A055573(k)

A260615 Irregular triangle read by rows: the n-th row is the continued fraction expansion of the sum of the reciprocals of the first n primes.

Original entry on oeis.org

0, 2, 0, 1, 5, 1, 30, 1, 5, 1, 2, 12, 1, 3, 1, 2, 1, 9, 1, 1, 7, 1, 2, 1, 9, 1, 2, 1, 2, 12, 7, 1, 2, 2, 13, 1, 1, 1, 8, 13, 5, 4, 1, 2, 5, 8, 1, 2, 6, 1, 1, 4, 10, 1, 2, 3, 1, 3, 1, 2, 238, 1, 28, 1, 42, 2, 2, 7, 1, 1, 4, 1, 1, 1, 6, 1, 41, 3, 1, 1, 51, 1, 9, 2, 3, 2, 5, 1, 2, 1, 6, 1, 1, 1, 3, 3, 3, 1, 1, 1, 3, 3, 1, 2, 19, 1, 13, 1, 1, 3, 4, 7, 1, 1, 3, 2, 1, 10

Offset: 1

Matthew Campbell, Aug 29 2015

			For row 3, the sum of the first three prime reciprocals equals 1/2 + 1/3 + 1/5 = 31/30. The continued fraction expansion of 31/30 is 1 + (1/30). Because of this, the terms in row 3 are 1 and 30.
From _Michael De Vlieger_, Aug 29 2015: (Start)
Triangle begins:
0,  2
0,  1,   5
1, 30
1,  5,   1,  2, 12
1,  3,   1,  2,  1,  9,  1,  1,  7
1,  2,   1,  9,  1,  2,  1,  2, 12,  7
1,  2,   2, 13,  1,  1,  1,  8, 13,  5,  4
1,  2,   5,  8,  1,  2,  6,  1,  1,  4, 10,  1,  2,  3,  1,  3
1,  2, 238,  1, 28,  1, 42,  2,  2,  7,  1,  1,  4
...
(End)

Matthew Campbell, Table of n, a(n) for n = 1..116505 The first 225 rows are in the b-file.

Cf. A000040.
For the continued fractions of the harmonic numbers, see A100398.
For the numerator of the sum, see A024451.
For the denominator of the sum, see A002110.

seq(op(numtheory:-cfrac(s,'quotients')),s=ListTools:-PartialSums(map2(`/`,1,[seq(ithprime(i),i=1..20)]))); # Robert Israel, Sep 06 2015

Table[ContinuedFraction[Sum[1/Prime@k, {k, n}]], {n, 11}] // Flatten (* Michael De Vlieger, Aug 29 2015 *)

row(n) = contfrac(sum(k=1, n, 1/prime(k)));
tabf(nn) = for(n=1, nn, print(row(n))); \\ Michel Marcus, Sep 18 2015

A113123 Numerator of next-best approximation to harmonic numbers. a(n) = Numerator of (A055573(n)-1)th convergent of n-th harmonic number, Sum_{k=1..n} 1/k.

Original entry on oeis.org

0, 1, 2, 2, 16, 22, 70, 106, 1836, 2639, 14281, 21167, 167857, 87932, 169452, 923889, 3590229, 950596, 40366604, 23213361, 517630, 1391957, 160363133, 222528683, 10125035246, 4324958013, 81828906108, 71315450571, 4320297286472

Offset: 1

Leroy Quet, Oct 14 2005

A100398 gives terms of continued fractions of harmonic numbers.

For n >= 2, a(n) = the denominator of the ratio equal to the continued fraction made by reversing the order of the terms of the continued fraction of the n-th harmonic number. (The numerator of this ratio is the numerator of the n-th harmonic number, A001008(n).) - Leroy Quet, Dec 24 2006

			H(6) = 49/20 = 2 +1/(2 +1/(4 +1/2)), so a(6) = numerator of 2 +1/(2 +1/4) = 22/9.

Cf. A100398, A055573, A113124.

More terms from Joshua Zucker, May 08 2006

A113124 Denominator of next-best approximation to harmonic numbers. a(n) = Denominator of (A055573(n)-1)th convergent of n-th harmonic number, Sum_{k=1..n} 1/k.

Original entry on oeis.org

1, 1, 1, 1, 7, 9, 27, 39, 649, 901, 4729, 6821, 52783, 27043, 51067, 273281, 1043807, 271979, 11378119, 6452207, 141997, 377141, 42943389, 58933037, 2653340203, 1122077597, 21027833867, 18159496967, 1090528730477, 236529224117

Offset: 1

Leroy Quet, Oct 14 2005

A100398 gives terms of continued fractions of harmonic numbers.

			H(6) = 49/20 = 2 +1/(2 +1/(4 +1/2)), so a(6) = denominator of 2 +1/(2 +1/4) = 22/9.

Cf. A100398, A055573, A113123.

More terms from Joshua Zucker, May 08 2006

A139036 a(n) = the number of 1's in the continued fraction expansion of the n-th harmonic number, H(n) = Sum_{k=1 to n} 1/k.

Original entry on oeis.org

1, 1, 2, 0, 2, 0, 2, 3, 5, 3, 1, 4, 6, 2, 3, 8, 8, 5, 8, 4, 10, 8, 8, 8, 7, 12, 9, 10, 13, 9, 8, 5, 10, 9, 12, 17, 15, 7, 9, 13, 8, 14, 12, 13, 14, 12, 11, 18, 17, 21, 19, 11, 12, 18, 16, 21, 33, 28, 19, 14, 20, 31, 19, 17, 21, 21, 16, 28, 23, 19, 18, 27, 40

Offset: 1

Leroy Quet, May 31 2008

nonn

			The 7th harmonic number is 1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 + 1/7 = 363/140, which has the continued fraction representation 2 + 1/(1 + 1/(1 + 1/(2 + 1/(5 + 1/5)))) = [2;1,1,2,5,5]. There are exactly two 1's in the continued fraction representation, so a(7) = 2.

Jinyuan Wang, Table of n, a(n) for n = 1..1000
Gonzalo Ciruelos, Python script that generates a(1)..a(n)

Cf. A100398.

Table[Count[ContinuedFraction[HarmonicNumber[n]],1],{n,100}] (* Harvey P. Dale, Nov 24 2016 *)

a(n) = #select(x->x==1, contfrac(sum(i=1, n, 1/i))); \\ Jinyuan Wang, Mar 01 2020

a(10)-a(15) from Gonzalo Ciruelos, Aug 02 2013

Corrected and extended by Harvey P. Dale, Nov 24 2016

cp's OEIS Frontend

A055573 Number of terms in simple continued fraction for n-th harmonic number H_n = Sum_{k=1..n} (1/k).

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A058027 Sum of terms of continued fraction for n-th harmonic number, 1 + 1/2 + 1/3 + ... + 1/n.

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A110020 Final term of the simple continued fraction for H(n), where H(n) = Sum_{k=1..n} 1/k.

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A112286 a(n) = numerator of sum of reciprocals of the terms of the continued fraction for H(n) = Sum_{k=1..n} 1/k.

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A112287 a(n) = denominator of sum of reciprocals of the terms of the continued fraction for H(n) = Sum_{k=1..n} 1/k.

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A139001 Partial sums of A055573 = number of terms in continued fraction of H(n)=sum(1/k,k=1..n).

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A260615 Irregular triangle read by rows: the n-th row is the continued fraction expansion of the sum of the reciprocals of the first n primes.

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A113123 Numerator of next-best approximation to harmonic numbers. a(n) = Numerator of (A055573(n)-1)th convergent of n-th harmonic number, Sum_{k=1..n} 1/k.

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